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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{1}{x^2-9}\]

$\begin{array}{1 1}\large \frac{1}{6} \log|x-3|-\large \frac{1}{6} \log|x+3| + c \\ \frac{1}{6} \log|x-3|+ \frac{1}{6} \log|x+3| + c \\ \frac{1}{6} \log|x+3|-\large \frac{1}{3} \log|x-3| + c \\ \frac{1}{6} \log|x+3|- \frac{1}{6} \log|x+3| + c \end{array} $

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1 Answer

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x-a)(x-b)},a\neq b\]
  • $(ii)\;$Form of the partial function\[\frac{A}{x-a}+\frac{B}{x-b}\]
  • $(iii)\;\int\frac{dx}{x-a}=log(x-a)+c.$
Given $I=\int\frac{dx}{(x^2-9)}=\int\frac{dx}{(x-3)(x+3)}.$
 
This can be resolved as\[\frac{A}{x-3}+\frac{B}{x+3}\]
 
$\Rightarrow 1=A(x+3)+B(x-3).$
 
Equating the coefficients of x we get,
 
0=A+B-------(1)
 
Equating the constant terms
 
1=3A+3B------(2)
 
To solve for A and B,multiply the equ(1) by 3 and subtract from equ(2)
 
3A+3B=1------(1)
3A+3B=0-----(2)
_______________
$6A=1\Rightarrow A=\frac{1}{6}.$
 
substituting for A in equ(1) we get $B=\frac{-1}{6}.$
 
Hence $A=\frac{1}{6}$ and $B=\frac{-1}{6}.$
 
Now substituting for A and B in I we get,
 
Therefore $I=\frac{1}{6}\int\frac{dx}{x-3}-\frac{1}{6}\frac{dx}{x+3}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{6}log|x-3|-\frac{1}{6}log|x+3|$
 
log|a|-log|b|=log|a/b|,similarly
 
$I=\frac{1}{6}log\frac{|x-3|}{|x+3|}+c.$

 

answered Feb 5, 2013 by sreemathi.v
edited Jul 21, 2013 by vijayalakshmi_ramakrishnans
 
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